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3 years ago

6

How to solve using perfect square x^2-4x-32=0

1 answer:

zmey [24]3 years ago

5 0

$Perfect\ square:(a-b)^2=a^2-2ab+b^2\\\\x^2-4x-32=0\ \ \ \ |add\ 32\ to\ both\ sides\\\\x^2-4x=32\\\\x^2-2x\cdot2=32\ \ \ \ |add\ 2^2\ to\ both\ sides\\\\x^2-2x\cdot2+2^2=32+2^2\\\\(x-2)^2=32+4\\\\(x-2)^2=36\iff x-2=-\sqrt{36}\ or\ x-2=\sqrt{36}\\\\x-2=-6\ or\ x-2=6\ \ \ \ |add\ 2\ to\ both\ sides\\\\\boxed{x=-4\ or\ x=8}$

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Answer:

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Step-by-step explanation:

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The answers to each of the given problems are;

1) Sum of two smallest integers = 23

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x + (x + 1) + (x + 2) + (x + 3) + 20 = 70

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x = 44/4

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2) Let the consecutive odd numbers be;

x, (x + 2) and (x + 4)

Sum of these consecutive odd numbers is;

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3) We are given the equation to find the acceleration as;

(v_final)² - (v_initial)² = 2ad

We are given;

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v_initial = 10 m/s

d = 100 m

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Read more about sum of integers at; brainly.com/question/17695139

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